Difference between revisions of "2009 AMC 12A Problems/Problem 22"

Latest revision as of 20:18, 13 August 2022

Problem

A regular octahedron has side length $1$ . A plane parallel to two of its opposite faces cuts the octahedron into the two congruent solids. The polygon formed by the intersection of the plane and the octahedron has area $\frac {a\sqrt {b}}{c}$ , where $a$ , $b$ , and $c$ are positive integers, $a$ and $c$ are relatively prime, and $b$ is not divisible by the square of any prime. What is $a + b + c$ ?

$\textbf{(A)}\ 10\qquad \textbf{(B)}\ 11\qquad \textbf{(C)}\ 12\qquad \textbf{(D)}\ 13\qquad \textbf{(E)}\ 14$

Solutions

Solution 1

Firstly, note that the intersection of the plane must be a hexagon. Consider the net of the octahedron. Notice that the hexagon becomes a line on the net. Also, notice that, given the parallel to the faces conditions, the line must be parallel to precisely $\frac{1}{3}$ of the sides of the net. Now, notice that, through symmetry, 2 of the hexagon's vertexes lie on the midpoint of the side of the "square" in the octahedron. In the net, the condition gives you that one of the intersections of the line with the net have to be on the midpoint of the side. However, if one is on the midpoint, because of the parallel conditions, all of the vertices are on the midpoint of a side. Thus, we have a regular hexagon with a side length of the midline of an equilateral triangle with side length 1, which is $\frac{1}{2}$ . Thus, the answer is $\frac {3\sqrt {3}}{8}$ , and $a + b + c = 14\ \mathbf{(E)}$ .

(Can somebody clarify this and provide a diagram?)

Solution 2

The plane cuts the octahedron into two congruent solids, which allows us to consider only the cut through the top half (a square pyramid). Because the cut is parallel to one side of the pyramid and must create two congruent solids, we can see that it must take the shape of a trapezoid with right, left, and top sides being 0.5 and a bottom side of 1. This is equivalent to a unit equilateral triangle with the tip cut off at the midpoint. Accordingly, the area of the polygon is $2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} \cdot \frac{3}{4}$ or $\frac{3\sqrt{3}}{8}$ , and $a + b + c = 14$ .

@@ Line 4: / Line 4: @@
 <math>\textbf{(A)}\ 10\qquad \textbf{(B)}\ 11\qquad \textbf{(C)}\ 12\qquad \textbf{(D)}\ 13\qquad \textbf{(E)}\ 14</math>
-== Solution ==
+== Solutions ==
-<asy><center>
+=== Solution 1 ===
-import three; currentprojection = orthographic(0.5,-3,1.4); pen g = rgb(0.8,1,0.8);
+Firstly, note that the intersection of the plane must be a hexagon. Consider the net of the octahedron. Notice that the hexagon becomes a line on the net. Also, notice that, given the parallel to the faces conditions, the line must be parallel to precisely <math>\frac{1}{3}</math> of the sides of the net. Now, notice that, through symmetry, 2 of the hexagon's vertexes lie on the midpoint of the side of the "square" in the octahedron. In the net, the condition gives you that one of the intersections of the line with the net have to be on the midpoint of the side. However, if one is on the midpoint, because of the parallel conditions, all of the vertices are on the midpoint of a side. Thus, we have a regular hexagon with a side length of the midline of an equilateral triangle with side length 1, which is <math>\frac{1}{2}</math>. Thus, the answer is<math> \frac {3\sqrt {3}}{8}</math>, and <math>a + b + c = 14\ \mathbf{(E)}</math>.
-triple[ P = {(1,0,0),(0,1,0),(-1,0,0),(0,-1,0),(0,0,1),(0,0,-1)]};
-void drawFrontFace(int x, int y, int z){ draw(P[x] -- P[y] -- P[z] -- cycle, linewidth(0.7)); }
+(Can somebody clarify this and provide a diagram?)
-void drawBackFace(int x, int y, int z){ draw(P[x] -- P[y] -- P[z] -- cycle, linetype("2 6")); }
-void fillFace(int x, int y, int z, pen c) {fill(P[x] -- P[y] -- P[z] -- cycle, c);}
-pair midpt(int x,int y){ return (P[x] + P[y])/2;}
-path planecut = midpt(1,0)--midpt(1,5)--midpt(2,5)--midpt(2,3)--midpt(4,3)--midpt(4,0)--cycle;
+=== Solution 2 ===
-fillFace(0,3,5,g);fillFace(1,2,4,g);fill(planecut,rgb(0.8,0.8,1));
+The plane cuts the octahedron into two congruent solids, which allows us to consider only the cut through the top half (a square pyramid). Because the cut is parallel to one side of the pyramid and must create two congruent solids, we can see that it must take the shape of a trapezoid with right, left, and top sides being 0.5 and a bottom side of 1. This is equivalent to a unit equilateral triangle with the tip cut off at the midpoint. Accordingly, the area of the polygon is <math>2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} \cdot \frac{3}{4}</math> or <math>\frac{3\sqrt{3}}{8}</math>, and <math>a + b + c = 14</math>.
-drawFrontFace(0,1,4);drawFrontFace(1,2,4);drawFrontFace(0,1,5);drawFrontFace(1,2,5);drawBackFace(2,3,4);drawBackFace(3,0,4);drawBackFace(2,3,5);drawBackFace(3,0,5);
-draw(planecut,linetype("4 4")+linewidth(0.7)); dot((0,0,0));
-</center></asy>
-If the plane divides the octahedron into two congruent solids, it goes through the center of the octahedron. As it is parallel to two opposite faces (colored above in green), it passes through the midpoints of the edges connecting the corresponding vertices of the faces. The distance between the center and any of the midpoints, as well as the distance between any consecutive midpoints, is found to be <math>1/2</math> (by midline and so forth). Thus, the intersection of the plane and the octahedron is a regular hexagon, and the answer is <math>6 \times \left(\frac {\left(\frac {1}{2}\right)^2 \sqrt {3}}{4}\right) = \frac {3\sqrt {3}}{8}</math>, and <math>a + b + c = 14\ \mathbf{(E)}</math>.
 == See also ==

2009 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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